You’re playing a game. The way it works is that there are a fixed number of predefined values “available” in the game. For instance, “0,100,200”. You always “own” exactly one of these values, and are randomly assigned one to start the game. Your objective is to end with as high a total as possible.
At any point you can “stand”. So obviously in this example if you have 200, you stand.
Your other option is to pay a fixed predefined cost in “points” (which is the same thing the value is adding up). If you do, you will switch to one of the other available values, at random.
So, let’s say that the fixed cost is 5. If you start with 100, you are paying 5 to randomly switch to either 0 or 200 (but effectively, -5 or 195, as you’ve paid 5). If you end up at 195, obviously you can’t do any better at this point. If you end up at -5, you can pay another 5 to random switch to 100 or 200, but now you’ve paid 5 twice, so 90 or 190.
The question is: when should you switch and when should you stand, given the possible values, your current value, and the cost to switch.
Now, the level 1 answer is that your EV for switching is (average of other values) - cost. If that is less than what you have, you should stand, otherwise you should switch. BUT, I’m pretty sure that is wrong, because of the guaranteed switching. By that logic, in this example, you would stand at 100, because the average of the other values is 100, and you have to pay 5 to switch, so your EV of switching is 95. BUT, crucially, if you end up at 0, then you get a REALLY good deal where you can pay another 5 to end up with an EV of 140. So I think that in this example, with these numbers, it is always correct to switch if the current value you own is not 200.
But, if instead of just (0,100,200) the numbers available were just (0,1,2,3…198,199,200) then I’m pretty sure this logic would not apply, and the straightforward EV math would be correct.
So… can anyone come up with a rule or formula that describe the optimal behavior? (If there is one, I don’t know it.)